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%\author{王立庆（2019级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数第六章：向量空间}
\title{第八章欧氏空间（8.1-8.2）考试解答 }
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\date{2023年5月10日}

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\begin{document}

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%\begin{abstract}
%%主要内容：
%7.3. 
%7.4. 
%7.5. 

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\begin{enumerate}

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\item %1
设 $V$ 是欧氏空间，设 $\alpha,\beta\in V$. 证明 
$\lvert \langle \alpha,\beta \rangle \rvert \le \lvert \alpha \rvert \cdot \lvert \beta \rvert$. 


\vspace{0.2cm}

{\color{red}解答：考虑实数 $t$ 的二次函数
\begin{eqnarray*}
f(t) &=& \langle t\alpha+\beta, t\alpha+\beta \rangle  \\ 
&=& t^2\langle \alpha,\alpha \rangle + 2t \langle \alpha,\beta \rangle + \langle \beta,\beta \rangle. 
\end{eqnarray*}
因为欧氏空间的内积是正定的，所以对任意实数 $t$, 总有 $f(t)\ge 0$. 所以判别式
\begin{eqnarray*}
\Delta=  4\langle \alpha,\beta \rangle ^2 - 4 \langle \alpha,\alpha \rangle \cdot \langle \beta,\beta \rangle \le 0. 
\end{eqnarray*}
开平方根得证。

}
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\item %2
考虑欧氏空间 $V=\mathbb{R}^3$ 中的向量组 
$%\begin{eqnarray*}
\Phi = \{ 
\alpha_1 = (2,0,1), \, 
\alpha_2 = (0,2,5), \,
\alpha_3 = (2,3,0)
\}.
$%\end{eqnarray*}

\begin{enumerate}
\item  证明向量组 $\Phi$ 是 $V$ 的一个基。
\item  用斯密特正交化方法，从 $\Phi$ 出发得到一个规范正交基。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  将这个向量组按列向量的方式排列成一个矩阵，计算它的行列式的值，可得 
$$ \begin{vmatrix} 2&0&2 \\  0&2&3 \\ 1&5&0 \\  \end{vmatrix} = -34 \neq 0. $$
因为行列式的值不为零，所以列向量组线性无关。所以这个列向量组是 $\mathbb{R}^3$ 的一个基。

\item  首先正交化，可得
\begin{align*}
\beta_1 &= \alpha_1 = (2,0,1), \\ 
\beta_2 &= \alpha_2 - \frac{\langle \alpha_2,\beta_1 \rangle}{\langle \beta_1,\beta_1 \rangle} \beta_1 
= (0,2,5) - \frac{5}{5} (2,0,1) = (-2,2,4),  \\ 
\beta_3 &= \alpha_3
 - \frac{\langle \alpha_3,\beta_1 \rangle}{\langle \beta_1,\beta_1 \rangle} \beta_1 
 - \frac{\langle \alpha_3,\beta_2 \rangle}{\langle \beta_2,\beta_2 \rangle} \beta_2 
 = (2,3,0) - \frac{4}{5}(2,0,1) - \frac{2}{24}(-2,2,4) = \frac{17}{30}(1,5,-2). 
\end{align*}
然后规范化，可得
\begin{align*}
\gamma_1 &= \frac{1}{\sqrt{5}} (2,0,1), \\ 
\gamma_2 &= \frac{1}{\sqrt{6}} (-1,1,2), \\ 
\gamma_3 &= \frac{1}{\sqrt{30}} (1,5,-2). 
\end{align*}
所求规范正交基为 $\Psi = \{\gamma_1, \gamma_2, \gamma_3\}$. 

\end{enumerate}

}

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\item %3
设 $V$ 是欧氏空间，设 $W$ 是 $V$ 的有限维子空间。设向量 $\alpha\in V$. 
\begin{enumerate}
\item  写出向量 $\alpha$ 在 $W$ 中的正射影的定义。 
\item  在 $\mathbb{R}^3$ 中，求向量 $\alpha=(1,2,3)$ 在由向量组 $\{(1,0,0), (0,1,0))\}$ 线性张成的子空间的正射影。
\item  在 $\mathbb{R}^4$ 中，求向量 $\alpha=(1,2,3,4)$ 在由向量组 $\{(1,1,1,0), (0,1,1,1))\}$ 线性张成的子空间的正射影。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  若存在 $\alpha_1$ 与 $\alpha_2$, 使得$\alpha = \alpha_1 + \alpha_2$ 以及 $\alpha_1\in W$ 且 $\alpha_2\in W^{\perp}$, 则称 $\alpha_1$ 是 $\alpha$ 在 $W$ 中的正射影。

\item  这个向量组
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\gamma_1 &=& (1,0,0) \\ 
\gamma_2 &=& (0,1,0)  
\end{array}\right. 
\end{eqnarray*}
已经是正交组，而且每个向量的长度等于1，所以所求正射影为
\begin{align*}
\alpha_1= \langle \alpha, \gamma_1 \rangle \gamma_1 + \langle \alpha, \gamma_2 \rangle \gamma_2 
= (1,2,0). 
\end{align*}
\item  
向量组
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\alpha_1 &=& (1,1,1,0) \\ 
\alpha_2 &=& (0,1,1,1)  
\end{array}\right. 
\end{eqnarray*}
不是正交组。所以先正交化，得到 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\beta_1 &=& \alpha_1 = (1,1,1,0), \\ 
\beta_2 &=& \alpha_2 - \frac{\langle \alpha_2,\beta_1 \rangle}{\langle \beta_1,\beta_1 \rangle} \beta_1 
= (0,1,1,1) - \frac{2}{3}(1,1,1,0) 
= \frac{1}{3}(-2,1,1,3).
\end{array}\right. 
\end{eqnarray*}
再规范化，得到
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\gamma_1 &=& \frac{1}{\sqrt{3}}(1,1,1,0), \\ 
\gamma_2 &=& \frac{1}{\sqrt{15}}(-2,1,1,3).   
\end{array}\right. 
\end{eqnarray*}
所以所求正射影为
\begin{align*}
\alpha_1= \langle \alpha, \gamma_1 \rangle \gamma_1 + \langle \alpha, \gamma_2 \rangle \gamma_2 
= \frac{6}{3}(1,1,1,0) + \frac{15}{15}(-2,1,1,3) 
= (0,3,3,3). 
\end{align*}


\end{enumerate}

}

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\end{enumerate}


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